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Mekhanik [1.2K]
3 years ago
15

What is 50% of 33 show work please​

Mathematics
1 answer:
olganol [36]3 years ago
8 0

Answer:

16.5

Step-by-step explanation:

50 x 33

-------------

100

= 16.5

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J. Reexamine the sequence 20, 14, 8, 2, ... from the problem
DENIUS [597]

Answer:

The n th of the given sequence is t_{n} = 26-6 n

Step-by-step explanation:

<u>Step 1</u> :-

Given sequence is 20,14,8,2,.......this sequence in arithmetic progression but this sequence is decreasing sequence.

given first term is 20 and difference isd = second term- first term = 14-20=-6

now the nth term of given sequence is

by using formula t_{n}=a+(n-1)d

t_{n}= 20+(n-1)(-6)

t_{n}= 20-6 n+6

final answer:-

t_{n} = 26-6 n

<u>verification</u>:-

t_{n} = 26-6 n

put n=1 we get first term is 20

put n=2 we get second term is 14

put n=3 we get third term is 8

put n=4 we get fourth term is 2

so the n th term of sequence is

t_{n} = 26-6 n

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3 years ago
I need help please?!!!
malfutka [58]

a<-28 I think. Can I have a Brainliest

6 0
3 years ago
Read 2 more answers
Write the words for the expression 3x(6-1)
fgiga [73]

Answer:

three <em>multipl</em><em>i</em><em>e</em><em>d</em><em> </em>by six <em>m</em><em>i</em><em>n</em><em>u</em><em>s</em><em> </em>one in <em>parentheses</em><em> </em>

Step-by-step explanation:

For the last part aka (6-1) just say that 6-1 is in parenthesis

Hope this helpss!! :DD♡

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Find the perimeter of the flag in the scale drawling. PLZ HELP I AM CONFUSED
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Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x.
Leni [432]

Answer:

a)

g(2.9) \approx -6.6

g(3.1) \approx -3.4

b)

The values are too small since g'' is positive for both values of x in. I'm speaking of the x values, 2.9 and 3.1.

Step-by-step explanation:

a)

The point-slope of a line is:

y-y_1=m(x-x_1)

where m is the slope and (x_1,y_1) is a point on that line.

We want to find the equation of the tangent line of the curve g at the point (3,-5) on g.

So we know (x_1,y_1)=(3,-5).

To find m, we must calculate the derivative of g at x=3:

m=g'(3)=(3)^2+7=9+7=16.

So the equation of the tangent line to curve g at (3,-5) is:

y-(-5)=16(x-3).

I'm going to solve this for y.

y-(-5)=16(x-3)

y+5=16(x-3)

Subtract 5 on both sides:

y=16(x-3)-5

What this means is for values x near x=3 is that:

g(x) \approx 16(x-3)-5.

Let's evaluate this approximation function for g(2.9).

g(2.9) \approx 16(2.9-3)-5

g(2.9) \approx 16(-.1)-5

g(2.9) \approx -1.6-5

g(2.9) \approx -6.6

Let's evaluate this approximation function for g(3.1).

g(3.1) \approx 16(3.1-3)-5

g(3.1) \approx 16(.1)-5

g(3.1) \approx 1.6-5

g(3.1) \approx -3.4

b) To determine if these are over approximations or under approximations I will require the second derivative.

If g'' is positive, then it leads to underestimation (since the curve is concave up at that number).

If g'' is negative, then it leads to overestimation (since the curve is concave down at that number).

g'(x)=x^2+7

g''(x)=2x+0

g''(x)=2x

2x is positive for x>0.

2x is negative for x.

That is, g''(2.9)>0 \text{ and } g''(3.1)>0.

So 2x is positive for both values of x which means that the values we found in part (a) are underestimations.

6 0
3 years ago
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